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i) inside surface area of the dome,

ii) volume of the air inside the dome.

Answer :

i)Given :

Cost of white washing the dome from inside = Rs. 498.96

Also, Cost of white washing 1 \({m}^2\) area = Rs. 2.00

Therefore, CSA of the inner side of dome

= Rs. \(\frac{498.96}{2} {m}^2 \)

= \( 249.48 {m}^2\)

ii) Let the inner radius of the hemispherical dome be r.

Now, CSA of inner side of dome = 249.48 \({m}^2\)

But, CSA = \(2{\pi}{r}^2\)

\(\Rightarrow \) 249.48 \({m}^2\) = \(2 × \frac{22}{7} × {r}^2\)

\(\Rightarrow \) \({r}^2 = (\frac{249.48 × 7}{2 × 22})\) \({m}^2\)

\(\Rightarrow \) r = 6.3 m.

Also, we have,

Volume of air inside the dome = Volume of hemispherical dome

= \(\frac{2}{3}{\pi}{r}^3\)

= \(\frac{2}{3} × \frac{22}{7} × {6.3}^3\) \({m}^3\)

= \(523.908\) \({m}^3\) (approx.)

Therefore, volume of the air inside the dome is \(523.908\) \({m}^3\).

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